googologywikiaorg-20200223-history
User blog:Ikosarakt1/Unusual number naming scheme
First, what is an "usual" number-naming scheme? We take a certain hierarchy, \(h_\alpha(n)\), and map a word for it up to certain n and \(\alpha\) (not necessarily do it for each \(\beta < \alpha\)). The restriction is only that the rule \(h_\alpha(n) = h_{\alphan}(n)\) always keeps. So, under that, Bowers' scheme is usual: \(h_1(100)\) = googol \(h_2(100)\) = giggol \(h_3(100)\) = gaggol ... \(h_\omega(100)\) = boogol \(h_{\omega*2}(100)\) = biggol (even though we have no word for \(h_{\omega+1}(100)\), it still works) \(h_{\omega*3}(100)\) = baggol ... \(h_{\omega^2}(100)\) = troogol Mixed-arrow notation Let's define so-called mixed-arrow notation as follows: # is an expression composed by\(\uparrow\)'s and\(\downarrow\)'s. Rule 1. (# is empty, no arrows at all) \(a\#b = ab = a*b\) Rule 2. (b=1, regardless of #) \(a\#1 = a\) Rule 3. (# ends at up-arrow) \(a\#\uparrow (b+1) = a\# (a\#\uparrow b)\) Rule 4. (# ends at down-arrow) \(a\#\downarrow (b+1) = (a\#\downarrow b) \# a\) Examples: \(3 \uparrow\downarrow 3 = (3 \uparrow 3) \uparrow 3 = 27 \uparrow 3 = 19683\) \(3 \downarrow\uparrow 3 = 3 \uparrow (3 \uparrow 3) = 3 \uparrow 27 = 7625597484987\) \(3 \uparrow\uparrow\downarrow 3 = (3 \uparrow\uparrow 3) \uparrow\uparrow 3 = (3 \uparrow\uparrow 3)^ \) \(3 \uparrow\downarrow\downarrow 3 = (3 \downarrow\downarrow 3) \downarrow\downarrow 3 = 19683 \downarrow\downarrow 3 = 19683^{19683^2}\) Note that \(a \uparrow \# b = a \downarrow \# b\) because they both drop to multiplication which is associative. Unusual naming scheme Now we want to form names for this unique notation. Let \(\downarrow\) be 0 and \(\uparrow\) be 1. Then we can map a binary expansion of each number to the arrow-expression. For example, 5 = 1012 = \(\uparrow\downarrow\uparrow\). So we can form the prefix for our names, adopting English numbers. For the suffix, I propose "-arrowal". For a and b we choose "3" as the smallest non-trivial case. So it forms the names for the numbers as follows: *Zerarrowal =\(3 \downarrow 3\) = 27 *Onarrowal =\(3 \uparrow 3\) = 27 (not greater than previous term) *Twarrowal =\(3 \uparrow\downarrow 3\) = 19683 *Thrarrowal =\(3 \uparrow\uparrow 3\) = 7625597484987 *Fourarrowal =\(3 \uparrow\downarrow\downarrow 3 = 19683^{19683^2}\) *Fivarrowal =\(3 \uparrow\downarrow\uparrow 3 = 3 \uparrow\downarrow (3 \uparrow\downarrow 3) = 3 \uparrow\downarrow 19683 = 3^{3^{19682}}\) *Sixarrowal =\(3 \uparrow\uparrow\downarrow 3 = (3 \uparrow\uparrow 3) \uparrow\uparrow 3 = 7625597484987 \uparrow\uparrow 3 = 7625597484987^{7625597484987^{7625597484987}}\) *Sevarrowal =\(3 \uparrow\uparrow\uparrow 3\) (exactly tritri in Bowers' system) *Eigarrowal =\(3 \uparrow\downarrow\downarrow\downarrow 3 = (3 \uparrow\downarrow\downarrow 3) \uparrow\downarrow\downarrow 3 = 19683^{19683^2} \uparrow\downarrow\downarrow 3 = 19683^{19683^2} \downarrow\downarrow\downarrow 3\) To estimate eigarrowal we have to determine how fast \(f(n) = n \downarrow\downarrow\downarrow 3\) grows. Consider it: \(n \downarrow\downarrow\downarrow 3 = (n \downarrow\downarrow n) \downarrow\downarrow 3 \approx (n \uparrow\uparrow 3) \uparrow\uparrow 3 \approx n \uparrow\uparrow 6\) . So eigarrowal is about \(19683^{19683^2} \uparrow\uparrow 6\), which is, as we know, much less than sevarrowal = \(3 \uparrow\uparrow 7625597484987\) . It turned out that in our naming scheme (n+1)-arrowal can be even lesser than n-arrowal. *Ninarrowal = \(3 \uparrow\downarrow\downarrow\uparrow 3\) *Tenarrowal = \(3 \uparrow\downarrow\uparrow\downarrow 3\) ... *Twentarrowal = \(3 \uparrow\downarrow\uparrow\downarrow\downarrow 3\) *Hundarrowal = \(3 \uparrow\uparrow\downarrow\downarrow\uparrow\downarrow\downarrow 3\) *Thousarrowal = \(3 \uparrow\uparrow\uparrow\uparrow\uparrow\downarrow\uparrow\downarrow\downarrow\downarrow 3 = 3 \uparrow^5 \downarrow\uparrow \downarrow^3\) (superscripts still can be used). I see no way to pin down this sequence to a standard ordinal hierarchy \(h_\alpha(n)\). Category:Blog posts